I've got a car with an internal volume of $12000$ $12000$ $L$ $L$ . If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from $1.0$ $1.0$ $a t m$ $atm$ to $1.4$ $1.4$ $a t m$ $atm$ ?

Question

13894 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-21T06:31:24

We use Boyle's Law: $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1 = P_2V_2$

Rearanging: $V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{1.0 \times 12000}{1.4} = 8571$ $V_2 = (P_1V_1)/P_2 =(1.0xx12000)/1.4 = 8571$ $L$ $L$

The answer is all above, but let me share a tip. It's possible to memorise Boyle's, Charles' and Gay-Lussac's laws and use them appropriately.

But I find it simpler to just memorise the Combined Gas Law, and then leave out anything that does not change in a particular question:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$

In this case, since temperature isn't mentioned, we simply leave it out and get Boyle's law.

I've got a car with an internal volume of 1200012000 LL. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from 1.01.0 atmatm to 1.41.4 atmatm?